Question

Prove the following identities

(1 + tan α tan β)² + (tan α – tan β)² = sec²α sec²β

Answer 1

LHS = (1 + tan α tan β)² + (tan α – tan β)²

= 1+ tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

= 1 + tan² α tan² β + tan² α + tan² β

= tan² α (tan² β + 1) + 1 (1 + tan² β)

= (1 + tan² β) (1 + tan² α)

We know that 1 + tan² θ = sec² θ

= sec² α sec² β

= RHS

Hence proved.